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ivyrianne's question at Yahoo! Answers regarding finding the height of a tree

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MarkFL

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Feb 24, 2012
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Here is the question:

A 40m high tower stands vertically on a hill side (sloping ground) which makes an angle of 18 degrees with the?

con't; horizontal. A tree also stands vertically up the hill from the tower. An observer on top of the tower finds the angle of depression from the top of the tree to be 26deg & the bottom of the tree to be 38deg. Find the height of the tree.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello ivyrianne,

I would begin by drawing a diagram:

ivyrianne.jpg

$T$ is the height of the tower.

$x$ is the horizontal distance between the tower and the tree.

$a$ is the vertical distance between the bottom of the tower and the bottom of the tree.

$b$ is the vertical distance between the top of the tower and the top of the tree.

$h$ is the height of the tree.

$\theta$ is the angle of inclination of the hill.

$\alpha$ is the angle of depression from the top of the tower to the top of the tree.

$\beta$ is the angle of depression from the top of the tower to the bottom of the tree.

Now, from the diagram, we observe the following relationships:

(1) \(\displaystyle a+h+b=T\)

(2) \(\displaystyle \tan(\theta)=\frac{a}{x}\)

(3) \(\displaystyle \tan(\alpha)=\frac{b}{x}\)

(4) \(\displaystyle \tan(\beta)=\frac{h+b}{x}\)

Solving (1) and (4) for $h+b$ and equating, we find:

\(\displaystyle T-a=x\tan(\beta)\)

Solving (2) for $a$, and substituting into the above, we may write:

\(\displaystyle T-x\tan(\theta)=x\tan(\beta)\)

Solving this for $x$, there results:

(5) \(\displaystyle x=\frac{T}{\tan(\beta)+\tan(\theta)}\)

Now, solving (4) for $h$, we get:

\(\displaystyle h=x\tan(\beta)-b\)

Solving (3) for $b$ and substituting into the above, we get:

\(\displaystyle h=x\tan(\beta)-x\tan(\alpha)\)

\(\displaystyle h=x\left(\tan(\beta)-\tan(\alpha) \right)\)

Substituting for $x$ from (5), we find:

\(\displaystyle h=\left(\frac{T}{\tan(\beta)+\tan(\theta)} \right)\left(\tan(\beta)-\tan(\alpha) \right)\)

\(\displaystyle h=\frac{T\left(\tan(\beta)-\tan(\alpha) \right)}{\tan(\beta)+\tan(\theta)}\)

Plugging in the given data:

\(\displaystyle \theta=18^{\circ},\,\alpha=26^{\circ},\,\beta=38^{\circ},\,T=40\text{ m}\)

we have:

\(\displaystyle h=\frac{(40\text{ m})\left(\tan\left(38^{\circ} \right)-\tan\left(26^{\circ} \right) \right)}{\tan\left(38^{\circ} \right)+\tan\left(18^{\circ} \right)}\approx10.6147758253\text{ m}\)